【算法训练】POJ1286 Necklace of Beads（Polya定理）

原创于 2018-08-08 19:07:20 发布 · 233 阅读

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本文探讨了在忽略旋转和反射重复的情况下，计算由红、蓝、绿三种颜色珠子组成的圆形项链的不同形式数量的问题。利用Polya定理，通过算法实现计算过程，给出具体代码实现。

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题意

Beads of red, blue or green colors are connected together into a circular necklace of n beads ( n < 24 ). If the repetitions that are produced by rotation around the center of the circular necklace or reflection to the axis of symmetry are all neglected, how many different forms of the necklace are there?
注意考虑对称翻转。

题解

Polya定理
这里写图片描述

代码

#include<cstdio>
#include<algorithm>
using namespace std;
typedef double db;
typedef long long ll;
typedef unsigned long long ull;
const int nmax = 1e6+7;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
const ull p = 67;
const ull MOD = 1610612741;
ll n,ans;
ll pow(ll t,ll b = 3ll){ return b * (t == 0?1ll:pow(t-1));}
int main(){
    while(scanf("%lld",&n)!=EOF && n !=-1){
        if(n == 0){ printf("0\n"); continue;}
        ans = 0;
        for(int i = 1;i<=n;++i)ans += pow(__gcd(n,1ll*i)-1);
        ans = ans / n;
        ans += (pow((n+1)/2-1) + pow(n/2+1-1))/2;
        printf("%lld\n",ans/2);
    }
    return 0;
}